Prove that $x^2+xy^2+xyz^2 ge 4xyz-4$ for positive real $x,y,z$

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Prove that $x^2+xy^2+xyz^2 ge 4xyz-4$ for postive real $x,y,z$.




I tried AM-GM but failed. I also can't apply Cauchy-Schwarz. I think we have to change this inequality to apply any well-known inequality. Please help me.






algebra-precalculus inequality







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    up vote
    2
    down vote

    favorite













    Prove that $x^2+xy^2+xyz^2 ge 4xyz-4$ for postive real $x,y,z$.




    I tried AM-GM but failed. I also can't apply Cauchy-Schwarz. I think we have to change this inequality to apply any well-known inequality. Please help me.






    algebra-precalculus inequality







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite












      Prove that $x^2+xy^2+xyz^2 ge 4xyz-4$ for postive real $x,y,z$.




      I tried AM-GM but failed. I also can't apply Cauchy-Schwarz. I think we have to change this inequality to apply any well-known inequality. Please help me.






      algebra-precalculus inequality







      share|cite|improve this question
















      Prove that $x^2+xy^2+xyz^2 ge 4xyz-4$ for postive real $x,y,z$.




      I tried AM-GM but failed. I also can't apply Cauchy-Schwarz. I think we have to change this inequality to apply any well-known inequality. Please help me.






      algebra-precalculus inequality




      algebra-precalculus inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 13 at 4:40









      max_zorn

      3,26061228




      3,26061228










      asked Sep 11 at 2:36









      Sufaid Saleel

      1,750628




      1,750628




















          2 Answers
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          accepted










          $$x^2+xy^2+xyz^2+4=x^2+2cdotdfracxy^22+4cdotdfracxyz^24+4$$



          Now using AM-GM inequality, $$dfracx^2+2cdotdfracxy^22+4cdotdfracxyz^24+41+2+4+1gesqrt[1+2+4+1]x^2cdotleft(cdotdfracxy^22right)^2cdotleft(dfracxyz^24right)^4cdot4$$



          Here is how I've identified the coefficients:



          let $$x^2+xy^2+xyz^2+4=acdotdfracx^2a+bcdotdfracxy^2b+ccdotdfracxyz^2c+dcdotdfrac4d$$



          Now by AM-GM, $$dfracacdotdfracx^2a+bcdotdfracxy^2b+ccdotdfracxyz^2c+dcdotdfrac4da+b+c+dge?$$



          Compare the exponents of $x,y,z$ to find



          $$a=d,b=2a,c=4a$$



          Choose $a=1$






          share|cite|improve this answer





























            up vote
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            $$x^2+xy^2+xyz^2-4xyz+4 = x^2+xy^2-4xy+4+xy(z-2)^2 = $$ $$x^2+x(y-2)^2+xy(z-2)^2-4x+4 =(x-2)^2+x(y-2)^2+xy(z-2)^2geq 0.$$






            share|cite|improve this answer


















            • 1




              Shouldn't we have $x^2 + x(y - 2)^2 + colorredxy(z - 2)^2 - 4x + 4 = (x - 2)^2 + x(y - 2)^2 + xy(z - 2)^2$?
              – N. F. Taussig
              Sep 11 at 8:22










            Your Answer





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            2 Answers
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            2 Answers
            2






            active

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            active

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            up vote
            4
            down vote



            accepted










            $$x^2+xy^2+xyz^2+4=x^2+2cdotdfracxy^22+4cdotdfracxyz^24+4$$



            Now using AM-GM inequality, $$dfracx^2+2cdotdfracxy^22+4cdotdfracxyz^24+41+2+4+1gesqrt[1+2+4+1]x^2cdotleft(cdotdfracxy^22right)^2cdotleft(dfracxyz^24right)^4cdot4$$



            Here is how I've identified the coefficients:



            let $$x^2+xy^2+xyz^2+4=acdotdfracx^2a+bcdotdfracxy^2b+ccdotdfracxyz^2c+dcdotdfrac4d$$



            Now by AM-GM, $$dfracacdotdfracx^2a+bcdotdfracxy^2b+ccdotdfracxyz^2c+dcdotdfrac4da+b+c+dge?$$



            Compare the exponents of $x,y,z$ to find



            $$a=d,b=2a,c=4a$$



            Choose $a=1$






            share|cite|improve this answer


























              up vote
              4
              down vote



              accepted










              $$x^2+xy^2+xyz^2+4=x^2+2cdotdfracxy^22+4cdotdfracxyz^24+4$$



              Now using AM-GM inequality, $$dfracx^2+2cdotdfracxy^22+4cdotdfracxyz^24+41+2+4+1gesqrt[1+2+4+1]x^2cdotleft(cdotdfracxy^22right)^2cdotleft(dfracxyz^24right)^4cdot4$$



              Here is how I've identified the coefficients:



              let $$x^2+xy^2+xyz^2+4=acdotdfracx^2a+bcdotdfracxy^2b+ccdotdfracxyz^2c+dcdotdfrac4d$$



              Now by AM-GM, $$dfracacdotdfracx^2a+bcdotdfracxy^2b+ccdotdfracxyz^2c+dcdotdfrac4da+b+c+dge?$$



              Compare the exponents of $x,y,z$ to find



              $$a=d,b=2a,c=4a$$



              Choose $a=1$






              share|cite|improve this answer
























                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted






                $$x^2+xy^2+xyz^2+4=x^2+2cdotdfracxy^22+4cdotdfracxyz^24+4$$



                Now using AM-GM inequality, $$dfracx^2+2cdotdfracxy^22+4cdotdfracxyz^24+41+2+4+1gesqrt[1+2+4+1]x^2cdotleft(cdotdfracxy^22right)^2cdotleft(dfracxyz^24right)^4cdot4$$



                Here is how I've identified the coefficients:



                let $$x^2+xy^2+xyz^2+4=acdotdfracx^2a+bcdotdfracxy^2b+ccdotdfracxyz^2c+dcdotdfrac4d$$



                Now by AM-GM, $$dfracacdotdfracx^2a+bcdotdfracxy^2b+ccdotdfracxyz^2c+dcdotdfrac4da+b+c+dge?$$



                Compare the exponents of $x,y,z$ to find



                $$a=d,b=2a,c=4a$$



                Choose $a=1$






                share|cite|improve this answer














                $$x^2+xy^2+xyz^2+4=x^2+2cdotdfracxy^22+4cdotdfracxyz^24+4$$



                Now using AM-GM inequality, $$dfracx^2+2cdotdfracxy^22+4cdotdfracxyz^24+41+2+4+1gesqrt[1+2+4+1]x^2cdotleft(cdotdfracxy^22right)^2cdotleft(dfracxyz^24right)^4cdot4$$



                Here is how I've identified the coefficients:



                let $$x^2+xy^2+xyz^2+4=acdotdfracx^2a+bcdotdfracxy^2b+ccdotdfracxyz^2c+dcdotdfrac4d$$



                Now by AM-GM, $$dfracacdotdfracx^2a+bcdotdfracxy^2b+ccdotdfracxyz^2c+dcdotdfrac4da+b+c+dge?$$



                Compare the exponents of $x,y,z$ to find



                $$a=d,b=2a,c=4a$$



                Choose $a=1$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Sep 11 at 5:39

























                answered Sep 11 at 5:34









                lab bhattacharjee

                218k14153268




                218k14153268




















                    up vote
                    6
                    down vote













                    $$x^2+xy^2+xyz^2-4xyz+4 = x^2+xy^2-4xy+4+xy(z-2)^2 = $$ $$x^2+x(y-2)^2+xy(z-2)^2-4x+4 =(x-2)^2+x(y-2)^2+xy(z-2)^2geq 0.$$






                    share|cite|improve this answer


















                    • 1




                      Shouldn't we have $x^2 + x(y - 2)^2 + colorredxy(z - 2)^2 - 4x + 4 = (x - 2)^2 + x(y - 2)^2 + xy(z - 2)^2$?
                      – N. F. Taussig
                      Sep 11 at 8:22














                    up vote
                    6
                    down vote













                    $$x^2+xy^2+xyz^2-4xyz+4 = x^2+xy^2-4xy+4+xy(z-2)^2 = $$ $$x^2+x(y-2)^2+xy(z-2)^2-4x+4 =(x-2)^2+x(y-2)^2+xy(z-2)^2geq 0.$$






                    share|cite|improve this answer


















                    • 1




                      Shouldn't we have $x^2 + x(y - 2)^2 + colorredxy(z - 2)^2 - 4x + 4 = (x - 2)^2 + x(y - 2)^2 + xy(z - 2)^2$?
                      – N. F. Taussig
                      Sep 11 at 8:22












                    up vote
                    6
                    down vote










                    up vote
                    6
                    down vote









                    $$x^2+xy^2+xyz^2-4xyz+4 = x^2+xy^2-4xy+4+xy(z-2)^2 = $$ $$x^2+x(y-2)^2+xy(z-2)^2-4x+4 =(x-2)^2+x(y-2)^2+xy(z-2)^2geq 0.$$






                    share|cite|improve this answer














                    $$x^2+xy^2+xyz^2-4xyz+4 = x^2+xy^2-4xy+4+xy(z-2)^2 = $$ $$x^2+x(y-2)^2+xy(z-2)^2-4x+4 =(x-2)^2+x(y-2)^2+xy(z-2)^2geq 0.$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Sep 11 at 13:13

























                    answered Sep 11 at 2:50









                    dezdichado

                    5,7251928




                    5,7251928







                    • 1




                      Shouldn't we have $x^2 + x(y - 2)^2 + colorredxy(z - 2)^2 - 4x + 4 = (x - 2)^2 + x(y - 2)^2 + xy(z - 2)^2$?
                      – N. F. Taussig
                      Sep 11 at 8:22












                    • 1




                      Shouldn't we have $x^2 + x(y - 2)^2 + colorredxy(z - 2)^2 - 4x + 4 = (x - 2)^2 + x(y - 2)^2 + xy(z - 2)^2$?
                      – N. F. Taussig
                      Sep 11 at 8:22







                    1




                    1




                    Shouldn't we have $x^2 + x(y - 2)^2 + colorredxy(z - 2)^2 - 4x + 4 = (x - 2)^2 + x(y - 2)^2 + xy(z - 2)^2$?
                    – N. F. Taussig
                    Sep 11 at 8:22




                    Shouldn't we have $x^2 + x(y - 2)^2 + colorredxy(z - 2)^2 - 4x + 4 = (x - 2)^2 + x(y - 2)^2 + xy(z - 2)^2$?
                    – N. F. Taussig
                    Sep 11 at 8:22

















                     

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